1967 IMO Problems/Problem 2
Contents
[hide]Problem
Prove that if one and only one edge of a tetrahedron is greater than , then its volume is .
Solution
Assume and let . Let be the feet of perpendicular from to and and from to , respectively.
Suppose . We have that , . We also have . So the volume of the tetrahedron is .
We want to prove that this value is at most , which is equivalent to . This is true because .
The above solution was posted and copyrighted by jgnr. The original thread can be found here: [1]
Remarks (added by pf02, September 2024)
The solution above is essentially correct, and it is nice, but it is so sloppily written that it borders the incomprehensible. Below I will give an edited version of it for the sake of completeness.
Then, I will give a second solution to the problem.
A few notes which may be of interest.
The condition that one side is greater than is not really necessary. The statement is true even if all sides are . What we need is that no more than one side is .
The upper limit of for the volume of the tetrahedron is actually reached. This will become clear from both solutions.
Solution
Assume that five of the edges are . Take them to be the edges other than . Denote . Let be the feet of perpendiculars from to , from to the plane , and from to , respectively.
At least one of the segments has to be . Suppose . (If were bigger that the argument would be the same.) We have that . By the same argument in we have . Since plane , we have , so .
The volume of the tetrahedron is
area of (height from .
We need to prove that . Some simple computations show that this is the same as . This is true because , and on the interval .
Note
is achieved when and all inequalities are equalities. This is the case when all sides except are , are the midpoint of , and (in which case the planes are perpendicular). In this case, , and as can be seen from an easy computation.
[This is an edited version of the solution by jgnr.]
Solution 2
Let be the set of tetrahedrons with five edges . This proof will show that there is a with one edge and such that , and that for any either or there is a finite sequence of tetrahedrons such that
.
The statement of the problem is a consequence of these facts.
We begin with two simple propositions.
Proposition
Let be a tetrahedron, and consider the transformations which rotate around while keeping fixed. We get a set of tetrahedrons, two of which, and are shown in the picture below. The lengths of all sides except are constant through this transformation.
1. Assume that the angles between the planes and , and and are both acute. If the perpendicular from to the plane is larger that the perpendicular from to the plane then the volume of is larger than the volume of .
2. Furthermore, the tetrahedron obtained when the position of is such that the planes and are perpendicular has the maximum volume of all tetrahedrons obtained from rotating around .
These statements are intuitively clear, since the volume of the tetrahedron is given by
area of height from .
A formal proof is very easy, and I will skip it.
Corollary
Given a tetrahedron , and an edge of it, we can find another tetrahedron such that , with an edge , and such that all the other edges of are equal to the corresponding edges of , the edge stretches between sides of which are perpendicular. When we chose a bigger , if we can choose . Or, we can choose such that it stretches between sides which are perpendicular.
(By "stretches between two sides" I mean that the end points of the edge are the vertices on the two sides which are not common to the two sides. In the picture above, stretches between the sides of .)
Lemma
Assume we have a tetrahedron with edges , such that . If there is an edge among then there is a tetrahedron with volume bigger than the volume of , whose edges are equal to those of , except for , which is replaced by an edge of size .
Proof
Case 1: If does not have any sides which are perpendicular, then the existence of follows from the corollary.
Case 2: Assume has exactly two sides which are perpendicular (like in the picture above). If and it were the only edge , then all the other edges are (because they were assumed to be ). In this case are equilateral with sides , and the planes can not be perpendicular since . (Indeed, an easy computation shows that if we take two equilateral triangles and place them perpendicular to each other, then .) So must be one of the other sides. Then again, the existence of follows from the corollary.
Case 3: Assume that three sides are perpendicular.
Assume the perpendicular sides are the ones meeting at , i.e. each pair of the planes meeting at are perpendicular. Since at least two of the edges it follows that (the sides of the right angle in a right triangle are less than the hypotenuse). Just for the sake of notation, assume . We can apply the corollary, and find a tetrahedron with volume bigger than the volume of , with edges equal to those of , except that is replaced by an edge .
Now the problem is very easy to prove. Let be a tetrahedron with edges , such that . Apply the lemma as many times as necessary (up to five times), successively replacing each edge by an edge . We obtain a tetrahedron with five edges , and one edge . If stretches between two perpendicular sides, we are done. If not, apply the corollary one more time to obtain a bigger tetrahedron in which is replaced by a larger edge which stretches between two perpendicular sides.
We obtain the same result as in the first solution: the largest tetrahedron is the one formed by two equilateral triangles with sides , having one side in common, with the two planes containing the triangles perpendicular. An easy calculation shows that the edge which is is in fact of length , and the volume of this tetrahedron is .
[Solution by pf02, September 2024]
See Also
1967 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |